Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1967 IMO Problems/Problem 2"

m
m
Line 63: Line 63:
 
<math>P, R</math> are midpoints of <math>AB</math>, and <math>Q = P</math> (in which case the planes
 
<math>P, R</math> are midpoints of <math>AB</math>, and <math>Q = P</math> (in which case the planes
 
<math>ABC, ABD</math> are perpendicular).  In this case, <math>CD = \frac{\sqrt{6}}{2}</math>,
 
<math>ABC, ABD</math> are perpendicular).  In this case, <math>CD = \frac{\sqrt{6}}{2}</math>,
−
as can be seen from an easy computation.
+
and <math>V = \frac{1}{8}</math> as can be seen from an easy computation.
  
 
(This is an edited version of the solution by jgnr.)
 
(This is an edited version of the solution by jgnr.)

Revision as of 13:32, 15 September 2024

Prove that if one and only one edge of a tetrahedron is greater than $1$, then its volume is $\le \frac{1}{8}$.

Solution

Assume $CD>1$ and let $AB=x$. Let $P,Q,R$ be the feet of perpendicular from $C$ to $AB$ and $\triangle ABD$ and from $D$ to $AB$, respectively.

Suppose $BP>PA$. We have that $CP=\sqrt{CB^2-BT^2}\le\sqrt{1-\frac{x^2}4}$, $CQ\le CP\le\sqrt{1-\frac{x^2}4}$. We also have $DQ^2\le\sqrt{1-\frac{x^2}4}$. So the volume of the tetrahedron is $\frac13\left(\frac12\cdot AB\cdot DR\right)CQ\le\frac{x}6\left(1-\frac{x^2}4\right)$.

We want to prove that this value is at most $\frac18$, which is equivalent to $(1-x)(3-x-x^2)\ge0$. This is true because $0<x\le 1$.

The above solution was posted and copyrighted by jgnr. The original thread can be found here: [1]


Remarks (added by pf02, September 2024)

The solution above is essentially correct, and it is nice, but it is so sloppily written that it borders the incomprehensible. Below I will give an edited version of it for the sake of completeness.

Then, I will give a second solution to the problem.

A few notes which may be of interest.

The condition that one side is greater than $1$ is not really necessary. The statement is true even if all sides are $\le 1$. What we need is that no more than one side is $> 1$.

The upper limit of $1/8$ for the volume of the tetrahedron is actually reached. This will become clear from both solutions.


Solution

Assume that five of the edges are $\le 1$. Take them to be the edges other than $CD$. Denote $AB = x$. Let $P, Q, R$ be the feet of perpendiculars from $C$ to $AB$, from $C$ to the plane $ABD$, and from $D$ to $AB$, respectively.

Prob 1967 2 fig1.png

At least one of the segments $AP, PB$ has to be $\ge \frac{x}{2}$. Suppose $PB \ge \frac{x}{2}$. (If $AP$ were bigger that $\frac{x}{2}$ the argument would be the same.) We have that $CP = \sqrt{BC^2 - PB^2} \le \sqrt{1 - \frac{x^2}{4}}$. By the same argument in $\triangle ABD$ we have $DR \le \sqrt{1 - \frac{x^2}{4}}$. Since $CQ \perp$ plane $ABD$, we have $CQ \le CP$, so $CQ \le \sqrt{1 - \frac{x^2}{4}}$.

The volume $V$ of the tetrahedron is

$V = \frac{1}{3} \cdot ($area of $\triangle ABD) \cdot$(height from $C) = \frac{1}{3} \cdot \left( \frac{1}{2} \cdot AB \cdot DR \right) \cdot CQ \le \left( \frac{1}{6} \cdot x \cdot \sqrt{1 - \frac{x^2}{4}} \cdot \sqrt{1 - \frac{x^2}{4}} \right) = \frac{x}{6} \left( 1 - \frac{x^2}{4} \right)$.

We need to prove that $\frac{x}{6} \left( 1 - \frac{x^2}{4} \right) \le \frac{1}{8}$. Some simple computations show that this is the same as $(1 - x)(3 - x - x^2) \ge 0$. This is true because $0 < x \le 1$, and $-x^2 - x + 3 > 0$ on the interval $(0, 1]$.

Note

$V = \frac{1}{8}$ is achieved when $x = 1$ and all inequalities are equalities. This is the case when all sides except $CD$ are $= 1$, $P, R$ are midpoints of $AB$, and $Q = P$ (in which case the planes $ABC, ABD$ are perpendicular). In this case, $CD = \frac{\sqrt{6}}{2}$, and $V = \frac{1}{8}$ as can be seen from an easy computation.

(This is an edited version of the solution by jgnr.)


Solution 2

Let $\mathcal{T}$ be the set of tetrahedrons with five edges $\le 1$. This proof will show that there is a $T \in \mathcal{T}$ with one edge $> 1$ and such that $\mathbf{volume} (T) = \frac{1}{8}$, and that for any $U \in \mathcal{T}$ either $U = T$ or there is a finite sequence of tetrahedrons $T_1, \dots, T_n$ such that

$\mathbf{volume} (U) = \mathbf{volume} (T_1) < \dots < \mathbf{volume} (T_n) = \mathbf{volume} (T)$.

The statement of the problem is a consequence of these facts.

We begin with two simple propositions.

Proposition

Let $ABCD$ be a tetrahedron, and consider the transformations which rotate $\triangle ABC$ around $AB$ while keeping $\triangle ABD$ fixed. We get a set of tetrahedrons, two of which, $ABC_1D$ and $ABC_2D$ are shown in the picture below. The lengths of all sides except $CD$ are constant through this transformation.

Prob 1967 2 fig2.png

1. Assume that the angles between the planes $ABD$ and $ABC$, and $ABD$ and $ABC_1$ are both acute. If the perpendicular from $C_1$ to the plane $ABD$ is larger that the perpendicular from $C$ to the plane $ABD$ then the volume of $ABC_1D$ is larger than the volume of $ABCD$.

2. Furthermore, the tetrahedron $ABC_2D$ obtained when the position of $C_2$ is such that the planes $ABD$ and $ABC_2$ are perpendicular has the maximum volume of all tetrahedrons obtained from rotating $\triangle ABC$ around $AB$.

These statements are intuitively clear, since the volume $V$ of the tetrahedron $ABCD$ is given by

$V = \frac{1}{3} \cdot ($area of $\triangle ABD) \cdot ($height from $C)$.

A formal proof is very easy, and I will skip it.

Corollary

Given a tetrahedron $T$, and an edge $e_1$ of it, we can find another tetrahedron $U$ such that $\mathbf{volume}(U) > \mathbf{volume}(T)$, with an edge $f_1 > e_1$, and such that all the other edges of $U$ are equal to the corresponding edges of $T$, $\mathbf{unless}$ the edge $e_1$ stretches between sides of $T$ which are perpendicular. When we chose a bigger $f_1$, if $e_1 < 1$ we can choose $f_1 = 1$. Or, we can choose $f_1$ such that it stretches between sides which are perpendicular.

(By "stretches between two sides" I mean that the end points of the edge are the vertices on the two sides which are not common to the two sides. In the picture above, $DC_2$ stretches between the sides $ABD, AC_2B$ of $ABC_2D$.)

Lemma

Assume we have a tetrahedron $T$ with edges $e_1, \dots, e_6$, such that $e_2, \dots, e_6 \le 1$. If there is an edge $e_m < 1$ among $e_2, \dots, e_6$ then there is a tetrahedron $U$ with volume bigger than the volume of $T$, whose edges are equal to those of $T$, except for $e_m$, which is replaced by an edge of size $1$.

Proof

Case 1: If $T$ does not have any sides which are perpendicular, then the existence of $U$ follows from the corollary.

Case 2: Assume $T$ has exactly two sides which are perpendicular (like $ABC_2D$ in the picture above). If $C_2D$ were the only edge $< 1$, then all the other edges are $= 1$ (because they were assumed to be $\le 1$). In this case $\triangle ABD, \triangle AC_2B$ are equilateral with sides $= 1$, and the planes can not be perpendicular since $C_2D < 1$. (Indeed, an easy computation shows that if we take two equilateral triangles $\triangle ABD, \triangle AC_2B$ and place them perpendicular to each other, then $CD = \frac{\sqrt{6}}{2} > 1$.) So $e_m$ must be one of the other sides. Then again, the existence of $U$ follows from the corollary.

Case 3: Assume that three sides are perpendicular.

Prob 1967 2 fig3.png

Assume the perpendicular sides are the ones meeting at $A$, i.e. each pair of the planes meeting at $A$ are perpendicular. Since at least two of the edges $BD, BC, CD \le 1$ it follows that $AB, AC, AD < 1$ (the sides of the right angle in a right triangle are less than the hypotenuse). Just for the sake of notation, assume $e_m = AB < 1$. We can apply the corollary, and find a tetrahedron $U$ with volume bigger than the volume of $T$, with edges equal to those of $T$, except that $AB$ is replaced by an edge $= 1$.

Now the problem is very easy to prove. Let $T$ be a tetrahedron with edges $e_1, \dots, e_6$, such that $e_2, \dots, e_6 \le 1$. Apply the lemma as many times as necessary (up to five times), successively replacing each edge $< 1$ by an edge $= 1$. We obtain a tetrahedron $U$ with five edges $f_2, \dots, f_6 = 1$, and one edge $f_1 = e_1$. If $f_1$ stretches between two perpendicular sides, we are done. If not, apply the corollary one more time to obtain a bigger tetrahedron in which $f_1$ is replaced by a larger edge which stretches between two perpendicular sides.

We obtain the same result as in the first solution: the largest tetrahedron is the one formed by two equilateral triangles with sides $= 1$, having one side in common, with the two planes containing the triangles perpendicular. An easy calculation shows that the edge which is $> 1$ is in fact of length $\frac{\sqrt{6}}{2}$, and the volume of this tetrahedron is $\frac{1}{8}$.

(Solution by pf02, September 2024)


See Also

1967 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1967_IMO_Problems/Problem_2&oldid=227261"